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I am confused as to how to take the total derivative $\frac{dKE}{dt}$, where $KE$ is the kinetic energy.

I know that $KE = 1/2 *m * \dot{\vec r} \cdot \dot{\vec r}$. From here, if I take derivative of both sides w.r.t. $dt$, how should I apply the chain rule and simplify?

Attempted Solution:

$$\frac{dKE}{dt} = \frac{\partial KE}{\partial t} \frac {dt}{dt}*\frac{\partial KE}{\partial {\vec r}} \frac {d{\vec r}}{dt}*\frac{\partial KE}{\partial \dot {\vec r}} \frac {d\dot {\vec r}}{dt}* ...$$ How do I proceed from here? The answer in most books is simply $m*\dot {\vec r}\cdot \ddot {\vec r}$, which I totally agree with if $KE$ is a function of $\dot {\vec r}$ only. Why should one assume that? Clearly, $\dot {\vec r}$ comes from ${\vec r}$ and $t$.

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  • $\begingroup$ Related: physics.stackexchange.com/q/9122/2451 $\endgroup$
    – Qmechanic ♦
    Commented Sep 19, 2019 at 16:33
  • $\begingroup$ @Qmechanic Thanks. I read the linked question and kind of understand what the answers mean, but when applied to a real example it fails. Check out my first comment, under the answer by Ekrem, for the example case. $\endgroup$
    – User 10482
    Commented Sep 19, 2019 at 18:05

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The equation you wrote is correct. All you need to do is take the partial derivative. I think this is where you got confused. You can check the difference between partial and total derivative. So here is the answer:

$\dfrac{\partial KE}{\partial t} =\dfrac{\partial ( \dfrac{1}{2}\cdot m \cdot \dot{\vec{r}} \cdot \dot{\vec{r}})}{\partial t} = 0 $

$\dfrac{\partial KE}{\partial {\vec{r}} } =\dfrac{\partial ( \dfrac{1}{2}\cdot m \cdot \dot{\vec{r}} \cdot \dot{\vec{r}})}{\partial \vec{r}} = 0 $

$\dfrac{\partial KE}{\partial {\dot{\vec{r}}} } =\dfrac{\partial ( \dfrac{1}{2}\cdot m \cdot \dot{\vec{r}} \cdot \dot{\vec{r}})}{\partial \dot{\vec{r}}} = m \cdot \dot{\vec{r} } $

Therefore, from what you wrote we can conclude that $ \dfrac{dKE}{dt} = m \cdot \dot{\vec{r}\cdot } \ddot{\vec{r}}$. Simply because there is no $t$ nor ${\vec{r}}$ dependence explicitly in KE function.

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  • $\begingroup$ Yes, the partial derivatives are confusing to me. Isn't $\dot{\vec r}$ a function of $t$ and $\vec {r}$ i.e. $\dot{\vec r}(t,\vec r)$? for example let $z = x+2y$ and $f(z) = 2z$. If I were to take $\frac{\partial f(z)}{\partial x}$, it would be zero with the "not explicitly dependent" argument, but clearly evaluates to $2$ if one substitutes for $z$ with $x+2y$ first. $\endgroup$
    – User 10482
    Commented Sep 19, 2019 at 17:06
  • $\begingroup$ I suggest you to look up this link: physics.stackexchange.com/questions/9122/… There are pretty good explanations about what you are asking. I don't think I can come up with a better one. $\endgroup$
    – Etg
    Commented Sep 19, 2019 at 17:21
  • $\begingroup$ I understand it in theory but when I try an actual example (like I gave in the first comment) it does not work. Can you explain using that (or similar) example? $\endgroup$
    – User 10482
    Commented Sep 19, 2019 at 18:00
  • $\begingroup$ Well, you can think as it is the definition of the partial derivative. In your example $f(z) = 2z$ is a function mapping $\mathbb{R} \rightarrow \mathbb{R}$. But $z$ also a function of $x$ and $y$ i.e. $z(x,y) = x + 2y$. If you substitute this function into $f(z)$, then you get another new function mapping $\mathbb{R}^2 \rightarrow \mathbb{R}$. Let's call this function$g(x,y)$. i.e. $g(x,y)=2x + 4y$. Therefore $\dfrac{\partial g(x,y)}{\partial x} = 2$ but $\dfrac{\partial f(z)}{\partial x} = 0 $. $\endgroup$
    – Etg
    Commented Sep 19, 2019 at 19:21

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